I’ve spent a goodly amount of time last week trying to craft a proof hinging on converting the infinite sum to an improper integral using the integrator , and comparing that one to those using the integrators and . But it doesn’t seem to be working. If you can make a go of it, I’ll be glad to hear it. Instead, here’s a proof adapted from Apostol.

We let be a positive decreasing function defined on some ray. For our purposes, let’s let it be , but we could use any other and adapt the proof accordingly. What we require in any case, though, is that the limit . We define three sequences:

First off, I assert that is nonincreasing, and sits between and . That is, we have the inequalities

To see this, first let’s write the integral defining as a sum of integrals over unit steps and notice that gives an upper bound to the size of on the interval . Thus we see:

From here we find that .

On the other hand, we see that . Reusing some pieces from before, we see that this is

which verifies that the sequence is decreasing. And it’s easy to check that , which completes our verification of these inequalities.

Now is a monotonically decreasing sequence, which is bounded below by , and so it must converge to some finite limit . This is the difference between the sum of the infinite series and the improper integral. Thus if either the sum or the integral converges, then the other one must as well.

We can actually do a little better, even, than simply showing that the sum and integral either both converge or both diverge. We can get some control on how fast the sequence converges to . Specifically, we have the inequalities , so the difference converges as fast as the function goes to zero.

To get here, we look back at the difference of two terms in the sequence:

So take this inequality for and add it to that for . We see then that . Then add the inequality for , and so on. At each step we find . So as goes to infinity, we get the asserted inequalities.

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